Schur's lemmas ↩

aliases: []
shorthands: {}
created: 2021-10-19 14:46:39
modified: 2022-01-13 14:56:28

Schur's lemma I

group, finite dimensional vector space and an irreducible representation.
If linear operator is such that then (i.e. a multiple of the identity).
Short: if a linear operator commutates with an irreducible representation, it can only be a multiple of the identity.

Conclusion

If we are unsure if is irreducible or not and we find an (not a multiple of the identity) that commutes with all the 's, then must be reducible.

Consequences

• Every Abelian (commutative) group has only 1D irreducible representations
• The fact that can be diagonalized globally is not surprising anymore
• Every irreducible representation has since it commutes with anything

Proof

1. has at least one eigenvector¹, .
2. Take a look at : is it also an eigenvector of ?
   because of commutativeness, so is also an eigenvector or with the same eigenvalue it is in the same eigensubspace of as .
3. Then the eigensubspace of is an invariant subspace of the representation as well². But is an irreducible representation, so it can only have one invariant subspace, that is the whole space . It can only commute with the identity matrix or a multiple of it.

Schur's lemma II

group, and vector spaces and irreducible representations.
If linear operator is such that then is either or invertible (then ).
Short: if and , then .

Proof

Let's take a look at and .

• 
  is an invariant subspace of and since is irreducible, or .
• .
  Then
  is an invariant subspace of and since is irreducible, or .

Now we have the following possibilities:

• means that is injective.
• means that .
• means that .
• means that is surjective.

So either or is injective and surjective bijective, invertible. Q.E.D.